What is the integral from 0 to 1 of 1/3 (x+y) ?

The integral from 0 to 1 of 1/3 (x+y) is 1/3 (x+1/2)

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integral from 0 to 1 of 1/3 (x+y)

\int_{0}^{1} \frac{1}{3} (x + y) d y

\frac{1}{3} (x + \frac{1}{2})

Solution steps

\int_{0}^{1} \frac{1}{3} (x + y) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{0}^{1} x + y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int_{0}^{1} x d y + \int_{0}^{1} y d y)

\int_{0}^{1} x d y = x

\int_{0}^{1} y d y = \frac{1}{2}

= \frac{1}{3} (x + \frac{1}{2})

\int_{0}^{1} - x^{3} - x^{2} + 2 x d x

\int_{0}^{1} e^{x - e^{x}} d x

\int_{\frac{1}{2}}^{\frac{3}{4}} 2 x d x

\int_{0}^{3} - 6 x d x

\int_{0}^{π} 4 x sin (2 x) d x

What is the integral from 0 to 1 of 1/3 (x+y) ?
The integral from 0 to 1 of 1/3 (x+y) is 1/3 (x+1/2)